# Thread: calculus optimize

1. ## calculus optimize

help me with this please (please indicate which one you are answering, A or B....both questions are not related to one another):

question A) a box is to be made frmo a square piece of cardboard that measures 12cm X 12cm by having equal squares cut from the corners of the sides. find the dimensions of the box of the greatest volume that can be made this way.

question B) a merchant ship traveling due north at 12km/h crosses the track of a second ship travelling due east at 9kn/h, at a time 100 min after the second ship passed that point. find the distance of the closest approach of the two ships.

2. Originally Posted by juiicycouture
question A) a box is to be made frmo a square piece of cardboard that measures 12cm X 12cm by having equal squares cut from the corners of the sides. find the dimensions of the box of the greatest volume that can be made this way.
See the diagram below.

Let the length of each side of the squares we cut out of the corners of the cardboard be x

Now volume = length * width * height

from the diagram, we see, length = width = 12 - 2x and the height is x

so V = x*(12 - 2x)^2
=> V = x(144 - 48x + 4x^2)
=> V = 144x - 48x^2 + 4x^3
for the max of this graph, we set V' = 0 (do you know why this is?)

now V' = 144 - 96x + 12x^2
set V'=0
=> 12x^2 - 96x + 144 = 0
=> x^2 - 8x + 12 = 0 ..................divided through by 12
=> (x - 6)(x - 2) = 0
=> x = 6, x = 2 .................which of these give the maximum? well we can do the second derivative test, but i'm not sure you know about that, so let's just plug in these values into the original equation and whichever is higher, that's the maximum.

when x = 6
V = 144(6) - 48(6)^2 + 4(6)^3 = 0

when x = 2
V = 144(2) - 48(2)^2 + 4(2)^3 = 224

so for maximum volume, the length of each side of the squares must be 2 cm

so the dimensions are length=width = 12 - 2(2) = 8, height = 2

3. Originally Posted by juiicycouture
help me with this please ...
question B) a merchant ship traveling due north at 12km/h crosses the track of a second ship travelling due east at 9kn/h, at a time 100 min after the second ship passed that point. find the distance of the closest approach of the two ships.

Hello,

1. I assume that there is a typo and that you mean a speed of 9 km/h (knots per hour are only plausible when you are knitting... )

2. Because you gave a time value in minutes I've converted all speed values into km/min:
12 km/h = 0.2 km/min
9 km/h = 0.15 km/min

3. Let t be the time. t = 0 means the the merchant ship MS is at the intercept of the 2 courses.

4. I use a coordinate system with the intercept as origin.

Then you get 2 values for the traveled distances w:

MS: w = 0.2*t (t in minutes!) and the second ship SS
SS: w = 0.15*t + 0.15*100 because the SS was 100 minutes earlier at the interception.

Now use the distance formula because the distance between the two ships is the hypotenuse of a right triangle (make a sketch of the situation!)

d² = (0.2*t)² + (0.15*t+15)²
d² = 0.04*t² + 0.0225*t² + 4.5*t + 225
d² = 0.0625*t² + 4.5*t + 225

If d has a minimum then d² has an extremum too. Therefore calculate the first derivative of d² and check if you found a minimum or a maximum afterwards.

(d²)' = 0.125*t + 4.5

(d²)' = 0 ===> 0.125*t + 4.5 = 0 ===> t = -36

That means 36 minutes befor the MS reaches the interception the distance has its smallest value.

d²(-36) = (0.2*(-36))² + (0.15*(-36)+15)² =
Code:
STOP! I have made a mistake.
You find the correct answer at Soroban's post.
So sorry!
93.6 km². Therefore the shortest distance is:

d = √(93.6 km²) ≈ 9.675 km

EB

4. Hello, juiicycouture!

I got a different answer for (B) . . .

(B) A merchant ship traveling due north at 12km/h crosses the track of a second ship
travelling due east at 9km/h, at a time 100 min after the second ship passed that point.
Find the distance of the closest approach of the two ships.
Code:
    B *
|     *
|           *   z
12t |                 *
|                       *
|                             *
* - - - - - - * - - - - - - - - - - *
O     15      P         9t          A

Ship A starts at point O and travels east for 100 minutes at 9 km/h.
. . It moves to point P: OP = 15 km.
In the next t hours, it travels 9t km to point A.

Ship B starts at point O travels north at 12 km/h.
In the next t hours, it travels 12t km to point B.

Let z = AB: . .= .(9t + 15)² + (12t)² .= .225t² + 270t + 225 .[1]

Then: .2z·z' .= .450t + 270 . . z' .= .(225t + 135)/z .= .0

Hence: .225t + 135 .= .0 . . t .= -0.6

The ships are closest 0.6 hours (36 minutes) before they crossed paths.

Substitute into [1]: . .= .225(-0.6)² + 270(-0.6) + 225 .= .144

. . Therefore: .z .= .12 km.